# Integration by substitution for sin/cos products

1. Sep 5, 2014

### phoneketchup

Ok so I might be doing something silly but I just don't understand what is going on here. So the integral:

i = ∫ sin x (cos x)^3 dx

First I say u = cos x. So du = - sin x dx.

So now I have i = ∫ - u^3 du. Which gives: i = -(1/4)u^4 or -(1/4)(cos x)^4. Easy.

But if I say u = sin x instead, this is what happens:

So du = cos x dx. And I say i = ∫ sin x (cos x)^2 cos x dx.

So I have i = ∫ u(1 - u^2) du or i = ∫ (u - u^3) du. WHAT.

Why am I getting two different answers?? Which one is right and why?

Thanks!

2. Sep 5, 2014

### HallsofIvy

Actually, it doesn't. It gives -(1/4)(cos x)^4+ C for any constant C.

You are not getting two different answers. The integral of "u- u^3" is (1/2)u^2- (1/4)u^4+ c which is equal to (1/2)(sin x)^2- (1/4)(sin x)^4+ c.

See what happens if you replace (sin x)^2 by 1- (cos x)^2 in that last formula.

3. Sep 5, 2014

### valerioperi

Actually you get the same answer. In fact when you put $u=\sin(x)$ you get $$\int u-u^3\,du=\frac{u^2}{2}-\frac{u^4}{4}+c=\frac{\sin(x)^2}{2}-\frac{\sin(x)^4}{4}+c$$
now, if you work on the first result you got, you'll see that: $$-\frac{\cos(x)^4}{4}+c=-\frac{(\cos(x)^2)^2}{4}+c=-\frac{(1-\sin(x)^2)^2}{4}+c=\frac{\sin(x)^2}{2}-\frac{\sin(x)^4}{4}+\frac{1}{4}+c$$
The two results are the same because it is an indefinite integral and they differ by an arbitrary constant. It could also have been solved without substitution noticing that $\cos(x)^2=1-\sin(x)^2$ and that $D[f(g(x)]=f'(g(x)) g'(x)$

Didn't realise we replied at the same time, sorry

4. Sep 6, 2014

### phoneketchup

Oops! I forgot to put the constant of integration. And yeah I see now.. it was pretty silly haha. Thanks! :)